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  1. twitter
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    @rajathsbhat

    • one year ago
  2. rajathsbhat
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    this one kinda trips you up at first glance. ok here's the thing- since the mag field is uniform, as you pull the loop, the flux doesn't change. no flux change = no e.m.f = no work done to pull the loop. Therefore, the answer is E.

    • one year ago
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    Wait hang on. If there is no flux change then the Work is constant right? Then the answer is A?

    • one year ago
  4. rajathsbhat
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    I'm very sorry. I was in a hurry and I read the question wrong. Neglect my previous reply. Here's the right analysis: Ok first of all, if there is no change in flux, it like having no magnetic field at all. You see my point? suppose you're pulling the ring through air without any mag field at all. But you are increasing it's velocity i.e. you are accelerating it. Since this is a graph of W versus v, let's see what the slope gives us, okay? \[\large \begin{align}\mathrm{d}W&=m \frac{\mathrm{d} v}{\mathrm{d} t}\times\mathrm{d} s\\ \frac{\mathrm{d}W}{\mathrm{d} v}&=m\frac{\mathrm{d} s}{\mathrm{d} t}\\ &=m\mathrm{d} v=\text{change in momentum.} \end{align}\] Now, m is constant. what is dv? dv is a very very small part of the velocity axis. You get it by cutting the velocity axis into an infinite number of infinitesimally small equal pieces. So that's a constant too. Therefore, the slope of this graph is constant. The only graph that shows this behavior is B.

    • one year ago
  5. rajathsbhat
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    The point you have to note here is that you don't have to do any work against the mag field.

    • one year ago
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    hang on a sec. when the loop partially out of the magnetic field, the loop experienced change in magnetic flux linkage, right? Work done need to be done to oppose that change.

    • one year ago
  7. rajathsbhat
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    but i thought the magnetic field extended on all sides indefinitely...

    • one year ago
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